# General Ring Theory/Ring extensions

Proposition (commutative ring extension is a module):

Let ${\displaystyle R}$ be a commutative ring and let ${\displaystyle S}$ be a commutative extension ring of ${\displaystyle R}$. Then ${\displaystyle S}$ with its own addition and the restriction of the multiplication of ${\displaystyle S}$ to ${\displaystyle R\times S}$ is a module over ${\displaystyle R}$.

Proof: From the axioms holding for rings, we deduce the module axioms as follows as follows: Let ${\displaystyle \lambda ,\mu \in R}$ and ${\displaystyle a,b\in S}$. Then

1. ${\displaystyle \lambda (a+b)=\lambda a+\lambda b}$ (distributivity)
2. ${\displaystyle (\lambda +\mu )a=\lambda a+\mu a}$ (distributivity)
3. ${\displaystyle \lambda (\mu a)=(\lambda \mu )a}$ (commutativity of multiplication)
4. ${\displaystyle 1a=a}$ (unit),

the ring axiom that's being used being indicated in the brackets. ${\displaystyle \Box }$

Proposition (Let ${\displaystyle S/R}$ be a ring extension. Then the function

${\displaystyle I\mapsto I\cap R}$

defines a function from ideals of ${\displaystyle S}$ to ideals of ${\displaystyle R}$.):

{{{2}}}

Proof: Indeed, ${\displaystyle I\cap R\leq R}$ because it is certainly closed under addition and multiplication by elements of ${\displaystyle R}$. ${\displaystyle \Box }$

Proposition (Let ${\displaystyle S/R}$ be a ring extension, and suppose that ${\displaystyle T\subseteq S}$ is a multiplicative set. Then ${\displaystyle T\cap R}$ is a multiplicative set of ${\displaystyle R}$.):

{{{2}}}

Proof: ${\displaystyle T\cap R}$ is closed under multiplication because both ${\displaystyle T}$ and ${\displaystyle S}$ are. ${\displaystyle \Box }$